Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif

#include <bits/stdc++.h>

using namespace std;

typedef long double ld;

#ifdef DEBUG
#define eprintf(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#else
#define eprintf(...) ;
#endif

#define sz(x) ((int) (x).size())
#define TASK "text"

const int inf = (int) 1.01e9;
const long long infll = (long long) 1.01e18;
const ld eps = 1e-9;
const ld pi = acos((ld) -1);

#ifdef DEBUG
mt19937 mrand(300); 
#else
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count()); 
#endif

int rnd(int x) {
  return mrand() % x;
}

void precalc() {
}

const int maxn = (int) 1e5 + 5;
int n, k;
int a[maxn], b[maxn];

bool read() {
  if (scanf("%d%d", &n, &k) < 2) {
    return false;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a[i], &b[i]);
    a[i] *= 2;
  }
  return true;
}

long long s[maxn];
int x[maxn];
int xc[maxn];
vector<int> xs;
long long dp[maxn];

int st[4 * maxn];

void buildst(int v, int tl, int tr) {
  st[v] = n;
  if (tl == tr - 1) {
    return;
  }
  int tm = (tl + tr) / 2;
  buildst(2 * v, tl, tm);
  buildst(2 * v + 1, tm, tr);
}

void update(int v, int tl, int tr, int l, int r, int val) {
  if (l == tl && r == tr) {
    st[v] = min(st[v], val);
    return;
  }
  int tm = (tl + tr) / 2;
  if (l < tm) {
    update(2 * v, tl, tm, l, min(r, tm), val);
  }
  if (r > tm) {
    update(2 * v + 1, tm, tr, max(l, tm), r, val);
  }
}

int get(int v, int tl, int tr, int pos) {
  int res = st[v];
  if (tl == tr - 1) {
    return res;
  }
  int tm = (tl + tr) / 2;
  if (pos < tm) {
    res = min(res, get(2 * v, tl, tm, pos));
  } else {
    res = min(res, get(2 * v + 1, tm, tr, pos));
  }
  return res;
}

void solve() {
  for (int i = 0; i < n; i++) {
    if (b[i] == 1 && a[i] > k) {
      printf("-1\n");
      return;
    }
  }
  s[0] = 0;
  for (int i = 0; i < n; i++) {
    s[i + 1] = s[i] + a[i];
  }
  x[0] = 0;
  for (int i = 0; i < n; i++) {
    x[i + 1] = ((x[i] - a[i]) % k + k) % k;
  }
  for (int i = 0; i <= n; i++) {
    xs.push_back(x[i]);
  }
  sort(xs.begin(), xs.end());
  xs.erase(unique(xs.begin(), xs.end()), xs.end());
  for (int i = 0; i <= n; i++) {
    xc[i] = lower_bound(xs.begin(), xs.end(), x[i]) - xs.begin();
  }
  buildst(1, 0, sz(xs));
  for (int i = n - 1; i >= 0; i--) {
    int j = get(1, 0, sz(xs), xc[i]);
    long long len = s[j] - s[i];
    if (j < n) {
      len = ((len + k - 1) / k) * k;
      dp[i] = len + dp[j];
    } else {
      dp[i] = len;
    }
    if (b[i] == 2) {
      continue;
    }
    int l = (k - a[i] + 1), r = 0;
    l = (l + x[i]) % k;
    r = (r + x[i]) % k;
    int lc = lower_bound(xs.begin(), xs.end(), l) - xs.begin();
    if (lc >= sz(xs)) {
      lc = 0;
    }
    int rc = lower_bound(xs.begin(), xs.end(), r) - xs.begin();
    if (rc >= sz(xs)) {
      rc = 0;
    }
    if (lc < rc) {
      update(1, 0, sz(xs), lc, rc, i);
    } else if (lc > rc) {
      update(1, 0, sz(xs), lc, sz(xs), i);
      update(1, 0, sz(xs), 0, rc, i);
    }
  }
  buildst(1, 0, sz(xs));
  long long res = infll;
  for (int i = 0; i < n; i++) {
    int j = get(1, 0, sz(xs), xc[i]);
    if (j >= n) {
      res = min(res, s[i] + dp[i]);
    }
    if (b[i] == 2) {
      continue;
    }
    int l = (k - a[i] + 1), r = 0;
    l = (l + x[i]) % k;
    r = (r + x[i]) % k;
    int lc = lower_bound(xs.begin(), xs.end(), l) - xs.begin();
    if (lc >= sz(xs)) {
      lc = 0;
    }
    int rc = lower_bound(xs.begin(), xs.end(), r) - xs.begin();
    if (rc >= sz(xs)) {
      rc = 0;
    }
    if (lc < rc) {
      update(1, 0, sz(xs), lc, rc, i);
    } else if (lc > rc) {
      update(1, 0, sz(xs), lc, sz(xs), i);
      update(1, 0, sz(xs), 0, rc, i);
    }
  }
  assert(res < infll);
  printf("%lld\n", res);
}

int main() {
  precalc();
#ifdef DEBUG
  assert(freopen(TASK ".in", "r", stdin));
  assert(freopen(TASK ".out", "w", stdout));
#endif
  while (read()) {
    solve();
#ifdef DEBUG
    eprintf("Time %.2f\n", (double) clock() / CLOCKS_PER_SEC);
#endif
  }
  return 0;
}